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Thread: Sin cos Formula

  1. #1

    Post Sin cos Formula

    Please help!! From this formula :[ sin(x+h) - sin (x) ] / h
    how do i get this : [ 2 sin h/2 cos (x + h/2) ] / h ???

  2. #2
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    Quote Originally Posted by silverfox12 View Post
    Please help!! From this formula :[ sin(x+h) - sin (x) ] / h
    how do i get this : [ 2 sin (h/2) cos (x + h/2) ] / h ???
    Use:

    Sin(A+B) = Sin(A)*Cos(B) + Cos(A)*Sin(B)

    Then

    Sin(A+B) - Sin(A-B) = ??

    replace A = x + h/2 & B = x - h/2

    and you are there......
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    Thanks man !

  4. #4
    I might actually need more help here . As you've said sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b)
    Using that : sin(x + h/2) - sin(x - h/2) = [sin(x)*cos(h/2) + cos(x)*sin(h/2)] - [sin(x)*cos(h/2) - cos(x)*sin(h/2)] if got it right . But after some work (namely just adding and subtracting) i get something like this cos(x)*sin(h/2) + cos(x)* sin(h/2)which i think equals 2*sin(h/2)*cos(x) the problem is that according to the textbook it should be like this 2*sin(h/2)* cos(x + h/2) could you please help with this too ? Any idea ?

  5. #5
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    Quote Originally Posted by silverfox12 View Post
    I might actually need more help here . As you've said sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b)
    Using that : sin(x + h/2) - sin(x - h/2) = [sin(x)*cos(h/2) + cos(x)*sin(h/2)] - [sin(x)*cos(h/2) - cos(x)*sin(h/2)] if got it right . But after some work (namely just adding and subtracting) i get something like this cos(x)*sin(h/2) + cos(x)* sin(h/2)which i think equals 2*sin(h/2)*cos(x) the problem is that according to the textbook it should be like this 2*sin(h/2)* cos(x + h/2) could you please help with this too ? Any idea ?
    Do the algebra carefully:

    sin(x+h) - sin(x) = sin(A+B) - sin(A-B) = 2 * sin(B) * cos(A) = 2 * sin(h/2) * cos(x+h/2)
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  6. #6
    Thanks for the answer and sorry for bothering but i'm really confused. Could you please explain what does A and B represent ? In the last post you wrote A =x + h/2 and B = x - h/2 but then in this one at the last equation you replaced B = h/2 ?

  7. #7
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    Do the algebra carefully (and using pencil + paper instead of staring at the screen) :

    sin(x+h) - sin(x)

    Substitute A = x + h/2 & B = h/2

    = sin(A+B) - sin(A-B)

    = 2 * sin(B) * cos(A)

    = 2 * sin(h/2) * cos(x+h/2)

    First post was an example - not to be used for spoon-feeding.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  8. #8
    The last question i swear .
    I understand this
    sin(a+b) - sin(a-b ) = 2cos(a)sin(b)

    But how do i determine what should i insert instead of A and B ?

    Like in this particular example sin(x + h) - sin( x) . You chose A = x + h/2 and B = h/2 but why ?

    sin(x + h/2) - sin(x - h/2)
    sin(A + B ) - sin(A - B )
    here shouldn't A be equal to x and B equal to h/2 given these formulas ??
    I'm getting something wrong just cant find out what exactly . But why do i chose A as the whole(x + h/2) and B only as the h/2 part of (x - h/2); What's the rule of choosing A and B ?

  9. #9
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    Quote Originally Posted by silverfox12 View Post
    The last question i swear .
    I understand this
    sin(a+b) - sin(a-b ) = 2cos(a)sin(b)

    But how do i determine what should i insert instead of A and B ?

    Like in this particular example sin(x + h) - sin( x) . You chose A = x + h/2 and B = h/2 but why ?

    sin(x + h/2) - sin(x - h/2)
    sin(A + B ) - sin(A - B )
    here shouldn't A be equal to x and B equal to h/2 given these formulas ??
    I'm getting something wrong just cant find out what exactly . But why do i chose A as the whole(x + h/2) and B only as the h/2 part of (x - h/2); What's the rule of choosing A and B ?
    You have the expression you are working with
    sin(x+h) - sin(x)
    You look to where you want to go
    sin(h/2) cos(x+h/2)

    The sin(s1) * cos(s2) looks likes a sin(a+b)[tex] \pm[/tex] sin(a-b). [Had it been a cos(s1) cos(s2) or sin(s1) sin(s2), we would use cos(a+b) [tex]\pm[/tex] cos(a-b)]. For convenience choose the minus sign and thus we have
    sin(a+b) - sin(a-b) = 2 sin(b) cos(a) = 2 sin(h/2) cos(x+h/2)
    So
    b = h/2
    and
    a = x+h/2
    Check:
    a+b = x + h
    a-b = x
    which is
    sin(x+h) - sin(x)
    and what we wanted.

  10. #10
    Thanks i understand now . (a+b) = x+ h and (a-b) = x
    so a = x + b and that we plug into (a+b) = x+h and we get 2b = h - > b = h/2 -> a = x +h/2
    Wish you good health for the help .

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